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12:45 AM
@Haran Did you see that I solved Peter´s question elementarily?
Which question
I gave two answers there, both simple.
Q: Can a number be equal to the sum of the squares of its prime divisors?

PeterIf $$n=p_1^{a_1}\cdots p_k^{a_k},$$ then define $$f(n):=p_1^2+\cdots+p_k^2$$ So, $f(n)$ is the sum of the squares of the prime divisors of $n$. For which natural numbers $n\ge 2$ do we have $f(n)=n$ ? It is clear that $f(n)=n$ is true for the square of any prime, but false for the other prim...

Looks like it can be easily bounded
How did you arrive at $c=d$?
Just read the proof with patience, there are no hard parts, also there is a comment which clarifies some steps.
1:12 AM
I´ve opened a new room guys, so when you feel that sometimes it is more appropriate to discuss there then you can discuss there, and when you feel that sometimes it is more appropriate to discuss here than you can discuss here. @MrPie @Mathphile @Peter @TheSimpliFire @Haran and all the others...
1 hour later…
2:38 AM
@MrPie Hey lad!
3 hours later…
5:52 AM
@Mathphile I'll just mention that The COnjecture Chatroom is close to 14 days of inactivity (after which it gets frozen).
I wasn't sure whether you want to keep the room or whether it is already abandoned.
@MartinSleziak Thanks for the information, I will suggest to @Peter and @Mathphile that we do more discussions in a room I opened a few hours ago, which is of quite a general character, instead of frequently opening special-purpose rooms. Do you think that is a good idea?
6:37 AM
Well, I know that many people object to keeping alive room related to one specific area - many of them with low activity. I mean the rooms like the ones mentioned here: List of chatrooms.
I still think that they might be useful. (Well, otherwise I would not put effort in keeping them alive.)
If nothing else, there are situations where several discussions are going on in the main chatroom at the same time - in such cases it might be useful to have a chatroom handy where to go with your conversation. And if it is related, for example, to linear algebra then posting in linear algebra room might increase chances that the discussion will be noticed.
Regarding Hlibert's Hotel, I suppose that it's intended mostly for a specific group of users. (Without concentrating on one specific area.)
in Hilbert's hotel, 6 hours ago, by Grešnik
room topic changed to Hilbert's hotel: A room for posing problems and for solving problems and also for general and specific discussions about mathematics, philosophy of mathematics, history of mathematics, logic, computational experiments, mathematical education and about some other themes. [computational-mathematics] [conjectures] [education] [logic] [math-history] [open-problem] [soft-question]
@Grešnik I think there'd be more people in that room if the main area was in, as you specify in the description, history/philosophy of mathematics, but then again there is already the SE site History of Science and Mathematics.
Otherwise there is too much overlap between this room, your room, Logic, etc.
BTW there is also Philosophy of Mathematics room. (It's a gallery room.)
@MartinSleziak though very inactive
On the Shoulders of Giants, which is the main chatroom on HSM, has very low activity.
@OmegaKrypton Our questions 1 and 2 are both on the HNQ!
6:48 AM
@MartinSleziak It´s not a problem at all if there are specialized rooms for various branches of mathematics as the ones that you have made and it seems that there would be more order if the questions suited for some room would be asked in that room and the idea behind Hilbert´s hotel is that users from some other rooms can participate there also if they feel that atmosphere is such that it is better to do some discussion there, it is about having more choices available at a single moment.
@Grešnik How are you planning to use your "Discussion between Peter and Gresnik" room?
Will that be replaced by your new room?
@TheSimpliFire Could be, that´s not my room, Peter made that room.
In a way, that would be a good idea, since these Discussion rooms are not as easily accessed by other users
Good an bad simultaneously, if you really want to know.
7:43 AM
@TheSimpliFire wow
2 hours later…
9:19 AM
@MartinSleziak thanks for notifying me
it's a good place to keep records
10:16 AM
@Grešnik At the moment, I am satisfied. And I am convinced that the proof is valid.
@Peter Did you see also the new one, simplified and more general?
By the way, I found a reason why the function $$f(n)=n^{n+1}+(n+1)^{n+2}$$ has a small factor for so many odd n.
I upvoted this great answer !
Are you interested in the reason for f(n) ?
You can elaborate if you want.
If $n$ is of the form $6k+1$ , then $f(n)$ is divisible by $3$
If $n$ is of the form $6k+3$ , then $f(n)$ is divisible by $n^2+n+1$
What about 6k+5
10:27 AM
Seems that no factors are forced in this case.
But : If $n+2$ is an odd prime , then $f(n)$ is divisible by $n+2$
@TheSimpliFire Hi
Hello @Peter
@TheSimpliFire I found forced factors for some forms of n for the above f(n), maybe there are more.
Is it without loss of generality that n is odd?
No, the even case is completely different.
10:34 AM
"brb" , what does this mean ?
Be right back.
1 hour later…
12:00 PM
hello @EuxhenH
hello @TheSimpliFire @MrPie distraction here ;)
Q: Grammy Winners Grading

Omega KryptonPrologue. I am bored. I have divided the Grammy winners in the past few years into the following tiers, according to a specific and objective criterion. How did I categorise them? The Classification - Tier 0 Swift, Smith Tier 1 Glover, Lipa, Hansen Tier 2 Musgraves, Hernand...

I think there are certain heuristics that support the idea of cancelling all primes since we are adding and removing sums of the form 6k-1 and 6k+1 @EuxhenH
The real difficulty in this is the unpredictable variation in k for each prime
@TheSimpliFire should i thank you?
@OmegaKrypton I was gonna say Tier 0: 1 vowel, Tier 1: 2 vowels and Tier 2: 3 vowels but some of the artists in Tier 2 break the pattern
12:06 PM
@TheSimpliFire lol i have an instinct @MrPie would get it this time
^ a hint
at least this time isn't part of the tags
I'll be back in ~10mins
12:18 PM
another conjecture
$\sum_{k=0}^n (-1)^k n^{n-k} $ is only prime for $n=2$
1:08 PM
hello @Haran
any ideas about this?
@Mathphile for odd $n$, $n^n+1=(n-1)(n^{n-1}-n^{n-2}+\cdots-1)=(n-1)\left[n^n-\sum\limits_{k=0}^n(-1)^k n^{n-k}\right]$
I might have made some mistakes in the signs
but you get the idea
Maybe bring the n^n to the LHS or something, I might have time later to work the full thing out
1:13 PM
@TheSimpliFire okay
@Mathphile For even $n$ , the above sum is disible by $n+1$ (the proof is not difficult) and for odd $n$, the above sum is divisble by $n-1$ (the proof is also not diffcult). Try to find it out.
@Peter okay thanks for the tip
@Mathphile Even better, the above sum is always divisible by $n+1$
1:33 PM
@Mathphile You can factor a 111-digit number, if you want.
dividing $$62^{63}+63^{64}$$
enter "siqs(<number>)"
@Peter i see how the sum is divisible for odd $n$ by $n-1$ but not $n+1$
@Peter can you show me how is it divisible by $n+1$ for odd case?
Hint : Every summand is congruent to $(-1)^n$ mod $(n+1)$
i see
@Peter using yafu?
The above factorization might take some hours. Yes, with yafu.
so siqs(62^63+63^64)
1:44 PM
No, input the number above
it is a factor of this number
I am not sure whether the new yafu version allows a good way to see the progress. Is the output normal or weird ?

starting SIQS on c111: 296458513247491613957940602520125679169924506500179548416412620499302880901240095492001218810908429181608669479

==== sieving in progress (1 thread): 227600 relations needed ====
==== Press ctrl-c to abort and save state ====
Bpoly 1051 of 4096: buffered 238 rels, checked 39539 (47.59 rels/sec)
Bpoly 2127 of 4096: buffered 462 rels, checked 80554 (46.18 rels/sec)
@Peter output
OK, still acceptable. You see that we need about 228k relations
I hope that the current number of relations can be seen from time to time.
At least the program works correctly.
@Peter it does seem so
1:53 PM
I think, tonight, when I am in the chessclub, the factorization will be finished.
what do relations mean here?
602 relations have been found
2:16 PM
The idea is to construct a nontrivial congruence $x^2\equiv y^2\mod n$ giving a non-trivial factor.
For this , many partial relations are needed.
2 hours later…
4:23 PM
Bpoly 2970 of 4096: buffered 617 rels, checked 113300 (41.11 rels/sec)
@Peter can you figure out from this how far we reached?
4:42 PM
@Mathphile A bad connection, here in the chess club. No, this line does not show us the progress.
6601 rels found: 6007 full + 594 from 382376 partial, ( 36.71 rels/sec)
we need "rels found" or something like that
yes, this is helpful.
so how far are we?
We need 228k , at the moment , we are not very far. But this will become better. However, it will still take some hours, perhaps until tomorrow, hard to estimate.
I would have expected more found relations
4:57 PM
@Mathphile 8k reached ?
7336 rels
7k from 228k, still a long way to go.
but we should see a speedup.
isn't ecm faster?
If we are lucky, yes. You can simultaneously run factor(<number>)
If the smallest factor has around 40 digits , it will probably be faster.
Could be worth a try ...
@Mathphile But we should keep siqs running, in the case ecm is without a result.
Did you start ?
not yet
i am rendering some shaders for my game atm
my cpu is at 100% @92 degrees C
5:12 PM
This is why siqs is so slow :)
no i just started the rendering
Do not take everything serious
although it does seem that siqs has slowed down quite a bit due to the rendering
is there any way to pause calculation in yafu?
you can in fact abort and restart it.
If you enter the same number, the calculation should continue where it stopped.
5:27 PM
OK, see you tomorrow. I am curious whether the result is there then.

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